Further Mathematics Questions and Answers

\(\begin{vmatrix} 3 & x \\ 2 & x - 2 \end{vmatrix} = 3(x - 2) - 2x = -2\)

\(3x - 6 - 2x = -2 \implies x - 6 = -2\)

\(x = -2 + 6 = 4\)

\(\int \frac{4}{x^{3}} \mathrm {d} x = \int 4x^{-3} \mathrm {d} x\)

\(\frac{4x^{-3 + 1}}{-2} = -2x^{-2}|_{1}^{2} = \frac{-2}{x^{2}}|_{1}^{2}\)

= \(\frac{-2}{2^{2}}  - \frac{-2}{1^{2}} = -\frac{1}{2}  + 2 = 1\frac{1}{2}\)

\(\tan \theta = \frac{m_{1} - m_{2}}{1 - m_{1}m_{2}}\)

\(m_{1} = \text{slope of 1st line } 4y = 3x + 5 \implies y = \frac{3}{4}x + \frac{5}{4}\)

\(m_{1} = \frac{3}{4}\)

\(m_{2} = \text{slope of 2nd line} 3y = 1 - 2x \implies y = \frac{1}{3} - \frac{2}{3}x\)

\(m_{2} = -\frac{2}{3}\)

\(\tan \theta = \frac{\frac{3}{4} - (-\frac{2}{3})}{1 - ((\frac{3}{4})(-\frac{2}{3}))} = \frac{\frac{17}{12}}{\frac{1}{2}}\)

\(\tan \theta = \frac{17}{6}\)

\(\theta \approxeq 70.6°\)

\(y = 2(2x + \sqrt{x})^{2}\)

Let \(u = 2x + \sqrt{x}\)

\(y = 2u^{2}\)

\(\frac{\mathrm d y}{\mathrm d u} = 4u\)

\(\frac{\mathrm d u}{\mathrm d x} = 2 + \frac{1}{2\sqrt{x}}\)

\(\therefore \frac{\mathrm d y}{\mathrm d x} = (\frac{\mathrm d y}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\)

= \(4u(2 + \frac{1}{2\sqrt{x}}) \)

= \(4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})\)

\(XY = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}\) is the distance between a point \(X(x_{1}, y_{1})\) and \(Y(x_{2}, y_{2})\).

\(XY = \sqrt{(3 - 5)^{2} + (5 - 1)^{2}} = \sqrt{20}\)

= \(2\sqrt{5} = 4. 467 \approxeq 4.5\)